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Prove that the area of the equilateral triangle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle.

Answers (1)

Let PQR is a right angle triangle which is a right angle at point R.

$PR = b, RQ = a$

Three equilateral triangles are drawn on the sides of triangle PQR that is PRS, RTQ and PUQ

Let $x_{1},x_{2}$ and $x_{3}$ are the areas of equilateral triangles respectively

To prove:- $x_{1}+x_{2}=x_{3}$

Using Pythagoras theorem in $\Delta PQR$ we get

$PQ^{2}+RQ^{2}+RP^{2}$

$PQ^{2}=a^{2}+b^{2}$

$PQ=\sqrt{a^{2}+b^{2}}$

The formula of area of an equilateral triangle is

$=\frac{\sqrt{3}}{4}\left ( side \right )^{2}$

$\therefore$ Area of equilateral $\Delta RTQ$

$x_{1}=\frac{\sqrt{3}}{4}\left ( a^{2} \right )\; \; \; \; \; \; \; \; ....(1)$

Area of equilateral $\Delta RSP$

$x_{2}=\frac{\sqrt{3}}{4} b^{2} \; \; \; \; \; \; \; \; ....(2)$

Area of equilateral $\Delta PQU$

$x_{3}=\frac{\sqrt{3}}{4}\left ( \sqrt{a^{2}+b^{2}} \right )^{2}$ $\left ( Q\; PQ=\sqrt{a^{2}+b^{2}} \right )$

$x_{3}=\frac{\sqrt{3}}{4} \left ( a^{2}+b^{2} \right )$

add equation (1) and (2) we get

$x_{1}+x_{2}=\frac{\sqrt{3}}{4}a^{2}+\frac{\sqrt{3}}{4}b^{2}$

$x_{1}+x_{2}=\frac{\sqrt{3}}{4}\left ( a^{2}+b^{2} \right )=x_{3}$

Hence $x_{1}+x_{2}=x_{3}$

Hence proved

Posted by

infoexpert23

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